Q. 42 - Calculus

Question

Evaluate each of the double integrals in Exercises 37–54 as iterated integrals.

$$\begin{gathered} \int {\int }_{R}x\sin x \cos y dA, \\ where R=\left\{\right(x,y\left)\right|-3\le x\le 2and-2\le y\le 2\} \end{gathered}$$

Step-by-Step Solution

Answer

The value is $-4\sin \left(2\right)\cos \left(2\right)+2{\sin }^{2}2 - 6\sin \left(2\right)\cos \left(3\right) + 2\sin \left(2\right)\sin \left(3\right)$

Step 1. Given Information :

Given double integrals :

$$\begin{gathered} \int {\int }_{R}x\sin x \cos y dA, \\ where R=\left\{\right(x,y\left)\right|-3\le x\le 2and-2\le y\le 2\} \end{gathered}$$

We want to evaluate each of the double integrals as iterated integrals.

Step 2. Solution:

$$\begin{gathered} U\sin g Fubini\text{'}s Theorem \\ \int {\int }_{R}x\sin x \cos y dA={\int }_{-2}^2{\int }_{-3}^{2}x\sin x \cos y dx dy \\ Evaluation procedure for the iterated integral we get \\ ={\int }_{-2}^2\left({\int }_{-3}^{2}x\sin x \cos y dx\right)dy \end{gathered}$$

$$\begin{gathered} By Fundamental Theorem of Calculus we have \\ ={\int }_{-2}^2\left(\cos y \left({[-x\cos x]}_{-3}^2+{[\sin x]}_{-3}^2\right)\right) dy \\ Evaluation of the inner antiderivative we get \\ ={\int }_{-2}^2\cos y \left[(-2 \cos 2 -3 \cos 3)+(\sin 2 + \sin 3)\right] dy \\ ={\int }_{-2}^2\cos y \left[(-2 \cos 2 +\sin 2 - 3 \cos 3 + \sin 3)\right] dy \end{gathered}$$

$$\begin{gathered} By Fundamental Theorem of Calculus we have \\ = \left[(-2 \cos 2 +\sin 2 - 3 \cos 3 + \sin 3)\right] {\left[\sin y\right]}_{-2}^2 \\ Evaluation of the outer antiderivative we get \\ = \left[(-2 \cos 2 +\sin 2 - 3 \cos 3 + \sin 3)\right] \left[\sin 2 + \sin 2\right] \\ = \left[(-2 \cos 2 +\sin 2 - 3 \cos 3 + \sin 3)\right] \left[2\sin 2\right] \\ = \left[-4\sin \left(2\right)\cos \left(2\right)+2{\sin }^{2}2 - 6\sin \left(2\right)\cos \left(3\right) + 2\sin \left(2\right)\sin \left(3\right)\right] \end{gathered}$$